Coin Change Total Number of Ways in DSA C#

Mental Model

We want to find how many different ways we can make a total amount using given coin values. Each coin can be used many times.

Analogy: Imagine you have different types of Lego blocks (coins) and you want to build a tower of a certain height (amount). You want to count all the different ways to stack these blocks to reach exactly that height.

coins: [1, 2, 3]
amount: 4

ways array (index = amount):
[1, 0, 0, 0, 0]
 ↑
Initially, ways[0] = 1 means there is 1 way to make amount 0 (use no coins).

Dry Run Walkthrough

Input: coins = [1, 2, 3], amount = 4

Goal: Find total number of ways to make amount 4 using coins 1, 2, and 3.

Step 1: Start with ways array: ways[0] = 1, others 0

ways: [1, 0, 0, 0, 0]

Why: We know there is exactly 1 way to make amount 0: use no coins.

Step 2: Use coin 1: update ways for amounts 1 to 4

ways: [1, 1, 1, 1, 1]

Why: For each amount, add ways[amount - 1] because coin 1 can make that amount by adding 1.

Step 3: Use coin 2: update ways for amounts 2 to 4

ways: [1, 1, 2, 2, 3]

Why: Add ways[amount - 2] to current ways to include combinations using coin 2.

Step 4: Use coin 3: update ways for amounts 3 to 4

ways: [1, 1, 2, 3, 4]

Why: Add ways[amount - 3] to include combinations using coin 3.

Result:

ways: [1, 1, 2, 3, 4]
Total ways to make 4 = 4

Annotated Code

DSA C#

using System;

class CoinChange {
    public static int TotalWays(int[] coins, int amount) {
        int[] ways = new int[amount + 1];
        ways[0] = 1; // base case: 1 way to make amount 0

        foreach (int coin in coins) {
            for (int a = coin; a <= amount; a++) {
                ways[a] += ways[a - coin]; // add ways to make (a - coin)
            }
        }

        return ways[amount];
    }

    static void Main() {
        int[] coins = {1, 2, 3};
        int amount = 4;
        int result = TotalWays(coins, amount);
        Console.WriteLine(result);
    }
}

ways[0] = 1; // base case: 1 way to make amount 0

Initialize ways array with 1 way to make amount 0 (no coins).

foreach (int coin in coins) {

Iterate over each coin to build up ways to make amounts.

for (int a = coin; a <= amount; a++) {

For each amount from coin value to target, update ways.

ways[a] += ways[a - coin]; // add ways to make (a - coin)

Add ways to make amount (a - coin) to ways[a] to include current coin.

Output

4

Complexity Analysis

Time: O(n * amount) because we process each coin for every amount up to target.

Space: O(amount) because we store ways counts for all amounts from 0 to target.

vs Alternative: Compared to recursive solutions without memoization, this dynamic programming approach avoids repeated calculations and is much faster.

Edge Cases

amount = 0

Returns 1 because there is exactly one way to make amount 0 (use no coins).

DSA C#

ways[0] = 1; // base case: 1 way to make amount 0

coins array empty

Returns 0 for any positive amount because no coins are available to make change.

DSA C#

foreach (int coin in coins) {

coins contain duplicates

Duplicate coin values are treated as separate denominations, leading to overcounting the number of ways.

DSA C#

foreach (int coin in coins) {

When to Use This Pattern

When asked to count the number of ways to make a total using unlimited coins, use the Coin Change Total Number of Ways pattern with dynamic programming to build solutions bottom-up.

Common Mistakes

Mistake: Using a 2D array and double counting combinations by order instead of combinations by coin sets.
Fix: Use a 1D array and process coins one by one to avoid counting permutations.

Summary

Counts how many ways to make a target amount using given coin denominations unlimited times.

Use when you need the total number of combinations, not just one solution or minimum coins.

The key is to build up the number of ways for each amount by adding ways from smaller amounts using each coin.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?

easy

A. The total sum of all coin denominations.

B. The minimum number of coins needed to make a target amount.

C. The maximum value of coins that can be used to reach the target.

D. The total number of ways to make a target amount using given coin denominations.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible combinations of coins that sum up to the target amount.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, here we count ways, not minimize coins.

3. Final Answer:

   The total number of ways to make a target amount using given coin denominations. -> Option D

4. Quick Check:

   Count ways, not minimum coins [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins needed
• Thinking about maximum coin value instead of count
• Ignoring that order of coins does not matter

2. Which of the following C# code snippets correctly initializes the DP array for the Coin Change Total Number of Ways problem for amount 5?

easy

A. int[] dp = new int[6]; dp[0] = 1;

B. int[] dp = new int[5]; dp[0] = 1;

C. int[] dp = new int[5]; dp[1] = 1;

D. int[] dp = new int[6]; dp[1] = 1;

Solution

1. Step 1: Determine DP array size

   For amount 5, dp array size should be 6 to include index 0 to 5.

2. Step 2: Initialize base case

   dp[0] = 1 means there is 1 way to make amount 0 (using no coins).

3. Final Answer:

   int[] dp = new int[6]; dp[0] = 1; -> Option A

4. Quick Check:

   Array size = amount+1, dp[0] = 1 [OK]

Hint: DP size = amount + 1; dp[0] = 1 base case [OK]

Common Mistakes:

• Using array size equal to amount instead of amount+1
• Initializing dp[1] instead of dp[0]
• Forgetting to set base case dp[0] = 1

3. What is the output of the following C# code snippet?

int[] coins = {1, 2};
int amount = 3;
int[] dp = new int[amount + 1];
dp[0] = 1;
foreach (int coin in coins) {
  for (int i = coin; i <= amount; i++) {
    dp[i] += dp[i - coin];
  }
}
Console.WriteLine(dp[amount]);

medium

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0] initially, dp[0]=1 means 1 way to make 0.

2. Step 2: Process coin 1

   For i=1 to 3: dp[1]+=dp[0]=1, dp[2]+=dp[1]=1, dp[3]+=dp[2]=1; dp=[1,1,1,1]

3. Step 3: Process coin 2

   For i=2 to 3: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2; dp=[1,1,2,2]

4. Final Answer:

   2 -> Option A

5. Quick Check:

   Ways to make 3 with coins 1,2 = 2 [OK]

Hint: Add dp[i-coin] for each coin and amount [OK]

Common Mistakes:

• Mixing up dp array updates order
• Forgetting to initialize dp[0] = 1
• Miscounting ways by double counting

4. Identify the error in the following C# code for Coin Change Total Number of Ways and select the fix.

int[] coins = {1, 2, 3};
int amount = 4;
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int i = 0; i <= amount; i++) {
  for (int coin : coins) {
    if (i >= coin) {
      dp[i] += dp[i - coin];
    }
  }
}
Console.WriteLine(dp[amount]);

medium

A. Change dp[0] = 1 to dp[1] = 1

B. Change outer loop to 'for (int i = 1; i <= amount; i++)' and inner loop to 'for (int coin in coins)'

C. Replace 'for (int coin : coins)' with 'foreach (int coin in coins)'

D. No error, code is correct

Solution

1. Step 1: Identify incorrect loop syntax

   C# does not support 'for (int coin : coins)', it uses 'foreach (int coin in coins)'.

2. Step 2: Correct the loop syntax

   Replace 'for (int coin : coins)' with 'foreach (int coin in coins)' to fix syntax error.

3. Final Answer:

   Replace 'for (int coin : coins)' with 'foreach (int coin in coins)' -> Option C

4. Quick Check:

   C# uses foreach for collections [OK]

Hint: Use foreach for arrays in C# [OK]

Common Mistakes:

• Using Java-style for-each syntax in C#
• Changing dp base case incorrectly
• Assuming code is correct without checking syntax

5. Given coins = {1, 3, 5} and amount = 7, how many total ways are there to make amount 7? Use the coin change total number of ways logic.

hard

A. 5

B. 4

C. 7

D. 6

Solution

1. Step 1: Initialize dp array

   dp size = 8, dp[0] = 1, dp = [1,0,0,0,0,0,0,0]

2. Step 2: Process coin 1

   Update dp for i=1 to 7: dp[i]+=dp[i-1]; dp becomes [1,1,1,1,1,1,1,1]

3. Step 3: Process coin 3

   For i=3 to 7: dp[3]+=dp[0]=2, dp[4]+=dp[1]=2, dp[5]+=dp[2]=2, dp[6]+=dp[3]=3, dp[7]+=dp[4]=3; dp=[1,1,1,2,2,2,3,3]

4. Step 4: Process coin 5

   For i=5 to 7: dp[5]+=dp[0]=3, dp[6]+=dp[1]=4, dp[7]+=dp[2]=4; dp=[1,1,1,2,2,3,4,4]

5. Final Answer:

   4 -> Option B

6. Quick Check:

   Ways to make 7 with coins 1,3,5 = 4 [OK]

Hint: Update dp for each coin from coin to amount [OK]

Common Mistakes:

• Miscounting ways by skipping coins
• Not updating dp array correctly
• Confusing total ways with minimum coins